Решите уравнение: cos2x=sinx

1 - 2sin^2x = sinx - 2sin^2x - sinx + 1 = 0  /*(-1) 2sin^2x + sinx - 1 = 0  sinx = t 2t^2 + t - 1 = 0  D = 1 + 4*2 = 9 = 3^2 t1 = ( - 1 + 3)/4 = 2/4 = 1/2 t2 = ( - 1  - 3)/4 = - 4/4 = - 1 1) sinx = 1/2 х1 = π/6 + 2πn,    n ∈ Z х2 = 5π/6 + 2πn,    n ∈ Z 2) sinx = - 1 x3 = - π/2 + 2πn, n ∈Z